Problem 11 - A First Course in Complex Analysis with Applications

Question

Use known results to expand the given function in a Maclaurin series. Give the radius of convergence \(R\) of each series. $$ f(z)=\sin z^{2} $$

Step-by-Step Solution

Answer

The series is \( z^2 - \frac{z^6}{3!} + \frac{z^{10}}{5!} - \ldots \) with \( R = \infty \).

Step 1: Recall the Maclaurin Series Formula for Sine

The Maclaurin series for a function is the Taylor series expansion at zero. For the function \( \sin x \), the Maclaurin series is given by \[sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \ldots\]

Step 2: Substitute \( z^2 \) for \( x \)

We need the Maclaurin series for \( \sin(z^2) \). Substitute \( z^2 \) for \( x \) into the Maclaurin series of \( \sin x \):\[sin(z^2) = z^2 - \frac{(z^2)^3}{3!} + \frac{(z^2)^5}{5!} - \frac{(z^2)^7}{7!} + \ldots\]This simplifies to:\[sin(z^2) = z^2 - \frac{z^6}{3!} + \frac{z^{10}}{5!} - \frac{z^{14}}{7!} + \ldots\]

Step 3: Determine the Radius of Convergence

The series for \( \sin(z^2) \) resembles the series for \( \sin z \), which is known to converge for all \( z \in \mathbb{C} \). Therefore, the same holds when substituting \( z^2 \). Thus, the radius of convergence \( R \) is infinite.

Key Concepts

Radius of Convergence

When exploring series, the **radius of convergence** is a crucial concept. It determines how far from the center of the series expansion a power series will converge. For the Maclaurin series of a function, the center is always at zero.
To find the radius of convergence, one typically uses the ratio test or root test. However, for well-known series like the Maclaurin series of sine, the findings can be directly applied.
In the case of the series for \(\sin z\), it converges for every complex number \(z\), indicating that the radius of convergence, \(R\), is infinite. This is because the sine function is entire, meaning it is smooth and has derivatives everywhere in the complex plane. Consequently, the same applies to \(\sin(z^2)\), since squaring \(z\) doesn’t affect the convergence of coefficients.

Complex Analysis

**Complex analysis** deals with functions that have complex numbers as variables. Complex numbers are numbers in the form \(a + bi\), where \(i\) is the imaginary unit \(\sqrt{-1}\).
In complex analysis, we often study the properties of functions like continuity, differentiability, and integration within the complex plane.

• One key feature is that entire functions, like exponential and trigonometric functions, are defined everywhere in the complex plane.
• These functions have series expansions that converge across all complex numbers.

The function \(\sin(z^2)\) falls into this category. Since both sine and squaring an entire function maintain the properties of being entire, \(\sin(z^2)\) has an infinite radius of convergence, making it smoothly varying over the entire complex plane.

Series Expansion

A **series expansion** is a way to represent functions using an infinite series of terms. The Maclaurin series is a special case of the Taylor series, centered at zero. For a function \( f(z) \), its Maclaurin series is:\[f(z) = f(0) + f'(0)z + \frac{f''(0)}{2!}z^2 + \frac{f'''(0)}{3!}z^3 + \ldots\]For \( \sin(z^2) \), the series is derived by substituting \( z^2 \) into the Maclaurin series of \( \sin x \). This substitution transforms each term's power:

• The term \( z \) transforms to \( z^2 \).
• The power increases accordingly, such as from \( x^3 \) to \( z^6 \), \( x^5 \) to \( z^{10} \), and so forward.

This series expansion:\[\sin(z^2) = z^2 - \frac{z^6}{3!} + \frac{z^{10}}{5!} - \frac{z^{14}}{7!} + \ldots\]allows evaluating \(\sin\) at any point in the complex plane using simple arithmetic involving these terms. Understanding this expansion grants insights into how functions behave locally and can approximate them with finite sums very effectively.